Consider a very long rectangular fin attached to a flat surface such that the temperature at the end of the fin is essentially that of the surrounding air, i. e. 20°C. Its width is 5.0 cm; thickness is 1.0 mm; thermal conductivity is 200 W/m·K; and base temperature is 40°C. The heat transfer coefficient is 20 W/m2 ·K. Estimate the fin temperature at a distance of 5.0 cm from the base and the rate of heat loss from the entire fin.

The fin temperature at a distance of 5.0 cm from the base and the rate of heat loss from the entire fin are;

T = 29.78°C and Q' = 2.856 W

We are given:

Length of fin; x = 5 cm = 0.05 m

Width of fin; w = 5 cm = 0.05 m

Thickness of fin; t = 1 mm = 0.001 m

Thermal conductivity; k = 200 W/mK

heat transfer coefficient; h= 20 W/m².K

Base temperature; T_b = 40°C

Ambient Temperature; T_∞ = 20°C

(T - T_∞)/(T_b - T_∞) = e^(-mx)

where m is gotten from the formula;

m = √(hp/KA)

where;

p is perimeter = 2w + 2t = 2(0.05) + 2(0.001) = 0.102 m

A is cross sectional area = wt = 0.001 * 0.05 = 0.00005 m²

Thus;

m = √(20 * 0.102/200 * 0.00005)

m = 14.28 /m

Thus;

(T - 20)/(40 - 20) = e^(-14.28*0.05)

Solving for T gives;

T = 29.78°C

Q' = [√(hpKA)]*(T_b - T_∞)

Q' = √(20 * 0.102 * 200 * 0.00005)]*(40 - 20)

Q' = 2.856 W

Read more about rate of heat loss at; link

attached below

Explanation: